## Tag `00M5`

Chapter 10: Commutative Algebra > Section 10.74: Tor groups and flatness

Lemma 10.74.8. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

- The module $M$ is flat over $R$.
- For all $i>0$ the functor $\text{Tor}_i^R(M, -)$ is zero.
- The functor $\text{Tor}_1^R(M, -)$ is zero.
- For all ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$.
- For all finitely generated ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$.

Proof.Suppose $M$ is flat. Let $N$ be an $R$-module. Let $F_\bullet$ be a free resolution of $N$. Then $F_\bullet \otimes_R M$ is a resolution of $N \otimes_R M$, by flatness of $M$. Hence all higher Tor groups vanish.It now suffices to show that the last condition implies that $M$ is flat. Let $I \subset R$ be an ideal. Consider the short exact sequence $0 \to I \to R \to R/I \to 0$. Apply Lemma 10.74.2. We get an exact sequence $$ \text{Tor}_1^R(M, R/I) \to M \otimes_R I \to M \otimes_R R \to M \otimes_R R/I \to 0 $$ Since obviously $M \otimes_R R = M$ we conclude that the last hypothesis implies that $M \otimes_R I \to M$ is injective for every finitely generated ideal $I$. Thus $M$ is flat by Lemma 10.38.5. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 17748–17760 (see updates for more information).

```
\begin{lemma}
\label{lemma-characterize-flat}
Let $R$ be a ring. Let $M$ be an $R$-module.
The following are equivalent:
\begin{enumerate}
\item The module $M$ is flat over $R$.
\item For all $i>0$ the functor $\text{Tor}_i^R(M, -)$ is zero.
\item The functor $\text{Tor}_1^R(M, -)$ is zero.
\item For all ideals $I \subset R$ we have $\text{Tor}_1^R(M, R/I) = 0$.
\item For all finitely generated ideals $I \subset R$ we have
$\text{Tor}_1^R(M, R/I) = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose $M$ is flat. Let $N$ be an $R$-module.
Let $F_\bullet$ be a free resolution of $N$.
Then $F_\bullet \otimes_R M$ is a resolution of $N \otimes_R M$,
by flatness of $M$. Hence all higher Tor groups vanish.
\medskip\noindent
It now suffices to show that the last condition implies that
$M$ is flat. Let $I \subset R$ be an ideal.
Consider the short exact sequence
$0 \to I \to R \to R/I \to 0$. Apply
Lemma \ref{lemma-long-exact-sequence-tor}. We get an
exact sequence
$$
\text{Tor}_1^R(M, R/I) \to
M \otimes_R I \to
M \otimes_R R \to
M \otimes_R R/I \to
0
$$
Since obviously $M \otimes_R R = M$ we conclude that the
last hypothesis implies that $M \otimes_R I \to M$ is
injective for every finitely generated ideal $I$.
Thus $M$ is flat by Lemma \ref{lemma-flat}.
\end{proof}
```

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